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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
Which of the following volume-temperature `(V-I)` plots represents the behaviour of `1 "mole"` of an ideal gas at the atmospheric pressure?A. B. C. D. |
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Answer» Correct Answer - C Volume of 1 mole of an ideal gas at 273 K and 1 atm is 22.4 L and that at 373 K and 1 atm pressure is calculated as, `V= (RT)/(P)=(0.0821xx373)/(1)=30.58L` |
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| 202. |
When the temperature is increased, surface tension of water:A. IncreaseB. DecreasesC. Remains constantD. Show irregular behaviour |
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Answer» Correct Answer - B Upon increase of temperature the internal energy of water of any system increases resulting in decrease in intermolecular force and surface tension. |
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| 203. |
Which of the following statements is not true?A. The ratio of the mean speed to the rms speed is independent of the temperature.B. The square of the mean speed of the molecuels is equal to the mean square speed at a certain temperatureC. Mean kinetic energy of the gas molecuels at any given temperature is independent of the mean sped.D. The difference between rms speed and mean speed at any temperature for different gases diminishes as larger and yet larger molar masses are considered. |
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Answer» Correct Answer - b |
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| 204. |
A rigid container contains 5 mole `H_(2)` gas at some pressure and temperature. The gas has been allowed to escape by simple process from the container due to which pressure of the gas becomes half of its initial pressure and temperature become `(2//3)^(rd)` of its initial. The mass of gas remaining is :A. 7.5 gB. 1.5 gC. 2.5 gD. 3.5 g |
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Answer» Correct Answer - a (a) `PV=(10)/(M)RT " "...(1)` Let x g of the gas remain in the flask when final pressure `(P)/(2),(2T)/(3)` `:. " " (P)/(2)V=((x)/(M))R((2T)/(3)) " "...(2)` `(1)//(2) implies2=(10)/(x)xx(3)/(2)` `x=(3xx10)/(4)=7.5 g` |
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| 205. |
If one mole of a monatomic gas `(gamma=5/3)` is mixed with one mole of a diatomic gas `(gamma=7/5),` the value of gamma for mixture isA. 1.4B. 1.5C. 1.53D. 3.07 |
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Answer» Correct Answer - b (b) `{:(C_(v)=(3)/(2)RT),(C_(p)=(5)/(2)RT):}` for monoatomic gas, `{:(C_(v)=(5)/(2)2RT),(C_(p)=(7)/(2)RT):}` for diatomic gas Thus, for mixture for 1 mole each, `C_(v)=((3)/(2)RT+(5)/(2)RT)/(2)` `and" "C_(p)=((5)/(2)RT+(7)/(2)RT)/(2)` Thus, `(C_(p))/(C_(v))=(3RT)/(2RT)=1.5` |
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| 206. |
The root mean square speed of 8 g of He is 300 `ms_(-1)`. Total kinetic energy of He gas is :A. 120 JB. 240 JC. 360 JD. None of these |
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Answer» Correct Answer - c (c) `300=sqrt((3RT)/(4xx10^(-3)))," "RT=120` Total K.E. of He gas`=(3)/(2)nRT` `=(3)/(2)xx(8)/(4)xx120J` `=360J` |
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| 207. |
The ratio of the average molecular kinetic energy of `UF_(6)` to that of `H_(2)`. Both at 300K is:A. `1:1`B. `7:2`C. `176:1`D. `2:7` |
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Answer» Correct Answer - a |
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| 208. |
Calculate the temperature at which the rms velocity of sulphur dioxide molecules is the same as that of oxygen at 300K.A. `600^(@)C`B. 600KC. 300KD. `300^(@)C` |
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Answer» Correct Answer - NA |
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| 209. |
Pick out correct statement among the following about the equal volume of `N_(2)(g)` and `O_(2)(g)` at 298K and 1 atm:A. The average translational kinetic energy per molecule is same of `N_(2)(g)` and `O_(2)(g)`B. The most probable speed of two gases is sameC. The toal translational kinetic energy of `N_(2)(g)` and `O_(2)(g)` is sameD. The absolute entropy of both gases is same. |
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Answer» Correct Answer - a |
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| 210. |
Select the correct statement :A. Internal energy of a real gas at a given temperature increases as the volume increasesB. Internal energy of an ideal gas at given temperature increase as the volume increasesC. Internal energy of an ideal gas molecules is not a function of temperatureD. The internal energy of a real gas at a constant temperature is independent of change in volume |
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Answer» Correct Answer - a (a) Internal energy of a substance is a function o ftemperature and volume (exept ideal gas). For ideal gas it is only function of temperature. |
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| 211. |
The average molecular velocity in a gas sample at 300K is 500m/s. The temperature of this gas is increased until the average velocity of its molecules is 1000m/s. what is the new temperature?A. 420KB. 573kC. 600kD. 1200k |
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Answer» Correct Answer - d |
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| 212. |
Statement-1: `CH_(4),CO_(2)` has value of Z (compressibility factor) less than one, generally. Statement-2: `Zlt1` is due to repulsive forces among the molecules.A. Statement-I is True, Statement-II is True : Statement-II is a correct explanation for Statement-IB. Statement-I is True, Statement-II is True : Statement-II is `NOT` a correct explanation for Statement-IC. Statement-I is True, Statement-II is False.D. Statement-I is False, Statement-II is True. |
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Answer» Correct Answer - c |
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| 213. |
The compressibility of a gas is less than unity at S.T.PA. `V_(m)gt22.7` litresB. `V_(m)gt22.7` litresC. `V_(m)=22.7` litresD. `V_(m)=45.4` litres |
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Answer» Correct Answer - c |
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| 214. |
At what temperature will most probable speed of the molecules of the second member of homologous series `C_(n)H_(2n-2)` be the same as that of `SO_(2)` at `527^(@)C`.A. `500^(@)C`B. `727^(@)C`C. `227^(@)C`D. None of these |
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Answer» Correct Answer - c (c) Second member is `C_(3)H_(4)`, `sqrt((2RT_(1))/(M_(1)))=sqrt((2RT_(2))/(M_(2)))` `T_(1)=T_(2)((M_(1))/(M_(2)))=800((40)/(64))K` `=500K " or "227^(@)C` |
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| 215. |
What will be the nature of forces at critical conditions for a real gas?A. AttractiveB. RepulsiveC. No net dominant forcesD. Depends on the types of the gas, attractive for most of them and repulsive for `H_(2)` and He. |
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Answer» Correct Answer - a |
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| 216. |
At what temperature will average speed of the molecules of the second member of the series `C_(n)H_(2n)` be the same to that of `Cl_(2)` at `627^(@)C`?A. 259.4 KB. 400 KC. 532.4 KD. None of these |
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Answer» Correct Answer - c (c) Second member of `C_(n)H_(2n)` series `=C_(3)H_(6)=42` `=sqrt((8RT_(1))/(piM_(1)))=sqrt((8RT_(2))/(piM_(2)))=(900)/(71)=(T_(2))/(42)` `T_(2)=532.4K` |
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| 217. |
If the `v_(rms)` is `30R^(1//2)` at `27^(@)C` then calculate the molar mass of gas in kilogram.A. 0.02 kg/molB. 0.001 kg/molC. 0.003 kg/molD. 1 kg/mol |
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Answer» Correct Answer - d (d) `U_("rms")=sqrt((3RT)/(M)` |
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| 218. |
Consider the following statements: The coefficient B in the virial equation of state (P) is independent of temperature (Q) is equal to zero at boyle temperature `PV_(m)=RT(1+(B)/(V_(m))+(C)/(V_(m)^(2))+......)` (R) has the dimension of molar volume which of the above statement are correct?A. P and QB. P and RC. Q andRD. P,Q and R |
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Answer» Correct Answer - c |
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| 219. |
Select the correct option(s) for an ideal gas.A. Most probable speed increases with increase in temperatureB. Fraction of particles moving with most probable speed increases with increase in temperatureC. Fraction of particles moving with most probable speed are more for `Cl_(2)` than `H_(2)` under similar conditions of T,P andVD. Most probable speed is more for `Cl_(2)` than `H_(2)` at same temperature. |
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Answer» Correct Answer - ac |
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| 220. |
Select the correct option(s) A. Pressure in container-I is 3 atm before opening the valve.B. Pressure after opening the valve is 3.57 atm.C. Moles in each compartment are same after opening the valve.D. Pressure in each compartment are same after opening the valve. |
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Answer» Correct Answer - ad |
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| 221. |
If `C` & `D` are the third & fourth virial coefficients. If `(D)/(C ) = (V_(c))/(x)`. Then find the value of `x`. |
| Answer» Correct Answer - 3 | |
| 222. |
The temperature at which the second virial coefficient of a real gas is zero is called . |
| Answer» Correct Answer - D | |
| 223. |
Find pressure (in atm) at point A, 10 cm above the bottom of container: A. `(106)/(76)`B. `(156)/(76)`C. `(101)/(76)`D. `(91)/(76)` |
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Answer» Correct Answer - c |
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| 224. |
One litre of `O_(2)` and one litre of `h_(2)` are taken in a vessel of 2 litre capacity at `NTP` The gases are made to combine to form water . The amount of gas left in vessel is .A. `1.7136 g`B. `0.7136 g`C. `0.8136 g`D. `1.8136 g` |
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Answer» Moles of `O_(2) "left" = (1 xx 0.5)/(0.0821 xx 273) = 2.23 xx 10^(-2)` `:.` Wt of `O_(2)` left `= 2.23 xx 10^(-2) xx 32 =0.7136 g` . |
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| 225. |
One litre of `O_(2)` and one litre of `h_(2)` are taken in a vessel of 2 litre capacity at `NTP` The gases are made to combine to form water . If the vessel is heated to `100^(@)C` the total pressure is .A. `0.03` atmB. `2.02 atm`C. `1.02 atm`D. `1.08 atm` |
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Answer» On heating to `100^(@)C,H_(2)O` exists as vapours `:.` Total moles pressent at `100^(@)C` = moles of `H_(2)O` +moles of `O_(2)` `=4.46 xx 10^(-2) +2.23 xx 10^(-2)` ` =6.69 xx 10^(-2)` Volume of vessel `=2 litre, T =373 K` `:. P = (nRT)/(V) = (6.69 xx 10^(-2) xx 0.0821 xx 373)/(2)` `= 1.02 atm` . |
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| 226. |
One litre of `O_(2)` and one litre of `h_(2)` are taken in a vessel of 2 litre capacity at `NTP` The gases are made to combine to form water . The weight of water formed .A. `7.03 xx 10^(-1) g`B. `8.03 xx 10^(-2) g`C. `8.03 xx 10^(-1) g`D. `7.03 xx 10^(-2) g` |
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Answer» At `NTP` the moles of `H_(2)O` formed =mole of `H_(2)` `=(PV)/(RT)=(1xx1)/(0.0821 xx273) =4.46 xx 10^(-2)` `:.` wt of `H_(2)O` formed ` =4.46 xx 10^(-2) xx18` ` = 8.03 xx 10^(-1) g` . |
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| 227. |
One litre of `O_(2)` and one litre of `h_(2)` are taken in a vessel of 2 litre capacity at `NTP` The gases are made to combine to form water . The mole of `O_(2)` used of formation of water is .A. `5.23 xx 10^(-2)`B. `4.23 xx 10^(-2)`C. `3.23 xx 10^(-2)`D. `2.23 xx 10^(-2)` |
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Answer» Volume of `O_(2)` used for formation of `H_(2)O=0.5` litre `:.` Moles of `O_(2)` used for foramation of `H_(2)O` at `NTP` ` = (0.5)/(22.4) =2.23 xx 10^(-2)` . |
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| 228. |
In the figure shown the pressure of the confined gas will be: A. 30 cm of HgB. 40 cm of HgC. 36 cm of HgD. 46 cm of Hg |
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Answer» Correct Answer - b |
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| 229. |
In the following arrangement find the pressure of the confined gas in cm of Hg |
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Answer» Correct Answer - `0026` |
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| 230. |
The drain cleaner Drainex contains small bits of aluminium which react with caustic soda to produce hydrogen What volume of hydrogen at `20^(@)C` aand one bar will be released when `0.15 g` of aluminium reacts ? . |
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Answer» `{:(,2A1+,2NaOH+2H_(2)Orarr,2NaA1O_(2),3H_(2)),("Initial moles",(0.15)/(27),-,-,-),("Final moles",-,-,-,(3)/(2)xx(0.15)/(27)):}` `:.` Moles of `H_(2) = (0.15 xx 3)/(2 xx 27) = 8.33 xx 10^(-3)` By `PV = nRT, P =1 "bar" = 0.987` atm `T =20 + 273 =293K` `0.987 xx V = 8.33 xx 10^(-3) xx 0.082 xx 293` `V = 0.2099 litre = 202.9 mL` . |
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| 231. |
The heat capacities of the diatomic molecules attain a limiting value at higher temperature At room temperature the translational and rotational degrees of freedom contribute to the heat capacities of gases while the vibrational degree of freedom becomes active only at higher temperature . |
| Answer» Explantion is correct reason for statement | |
| 232. |
Absolute temperature of diatomic has in increased eight fold, where it dissociates completely into atom. How many time will be the new rms velocity-compared to initial ? |
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Answer» Correct Answer - 4 |
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| 233. |
At room temperature the rms speed of the molecules of a certain diatomic gas is formed to be `1930m//sec` The gas is .A. `H_(2)`B. `F_(2)`C. `O_(2)`D. `CI_(2)` |
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Answer» `u_(rms) = sqrt((3RT)/(M))=sqrt((3xx8.314xx300)/(M))=1930` `:. M =2 xx 10^(-3) kg` . |
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| 234. |
Let the most probable velocity of hydrogen molecules at a temperature `t^(@)C` is `V_(0)`. Suppose all the molecules dissociates into atoms when temperature is raised to `(2t+273).^(@)C` then the new rms velocity isA. `sqrt((2)/(3))V_(0)`B. `sqrt(3(2+(273)/(t)))V_(0)`C. `2sqrt(3)V_(0)`D. `sqrt(6)V_(0)` |
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Answer» Correct Answer - d |
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| 235. |
A gas will approach ideal behaviour atA. Low temperature ideal low pressureB. Low temperature ideal low pressureC. High temperature ideal low pressureD. High temperature ideal low pressure |
| Answer» Real gases approach ideal gas behaviour at high temperature and low pressure . | |
| 236. |
A sample of He gas in a flexible container at room temperature exhibits a certain pressure. What will be the new pressure when the absolute temperature and volume of the container are both halve? The pressure of the He will be:A. the sameB. doubledC. halvedD. quadrupled |
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Answer» Correct Answer - a |
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| 237. |
If the absolute temperature of a sample of gas in a fixed volume container is quadrupled, then the root mean square speed in the initial state `u_(i)` and that in the final stage `u_(f)` be related as .A. `u_(f)=u_(i)//4`B. `u_(f)=u_(i)//2`C. `u_(f)= 2u_(i)`D. `u_(f)= 4u_(i)` |
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Answer» `u_(i) prop sqrtT_(1)` `u_(f) prop sqrtT_(2)` `(u_(i))/(u_(f)) = sqrt(T_(1)/(T_(2))) = sqrt(T_(1)/(4T_(1))) = (1)/(2)` . |
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| 238. |
The freezing point of `0.05M` urea solution is differnet from that of `0.05 M` sodium chloride solution The depression in freezing point is directly propotional to the number of species present in the solution . |
| Answer» Explantion is correct reason for statement | |
| 239. |
What will be the pressure of the gas mixture when 0.5 litre of `H_(2)` at 0.8 bar and 2.0 litre of oxygen at 0.7 bar are introduced in a 1 litre vessel at `27^(@)C` . |
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Answer» For `H_(2) n = (PV)/(RT) = (0.8 xx 0.5)/(RT) = (0.4)/(RT)` For `O_(2) n = (0.7 xx 2.0)/(RT) = (1.4)/(RT)` `:.` Total moles in mixture ` =(0.4)/(RT) + (1.4)/(RT) = (1.8)/(RT)` Using `PV = nRT` for 1 litre container Now `P_(mixture) xx 1 = (1.8)/(RT) xx RT = 1.8` bar |
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| 240. |
A vessel contains `14 g` of `H_(2)` and `96g` of `O_(2)` at `STP` Calculate (a) The volume of vessel (b) the pressure in the vessel when chemical reaction is induced in the vessel by passing electric spark till one of the gas is consumed The temperature brought back to `273 K` Neglect the volume of water formed . |
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Answer» `n_(H_2) = (14)/(2) =7, n_(O_2) = (96)/(32) =3` `:. PV = n_(T)RT` `1 xx V =(7+3) xx 0.0821 xx 273` `:. V =224.11` litre `{:(,H_(2)+,½O_(2)rarr,H_(2)O,),("mole of"=0,7,3,0,),("mole after reaction",(7-6),0,6,),(,=1,,,):}` Thus 1 mole of `H_(2)` is left `:. Pxx 224.11 xx =1 xx 0.0821 xx 273` `:. P_(H_(2)) = 0.10 atm` . |
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| 241. |
A gas diffuses one-thrid as fast as `O_(2)` at `100^(@)C`. This gas could be:A. He(M=4)B. `C_(2)H_(5)(M=48)`C. `C_(7)H_(12)(M=96)`D. `C_(5)F_(12)(M=288)` |
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Answer» Correct Answer - d |
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| 242. |
In two vessels of 1 litre each at athe same temperature 1g of `H_(2)` and 1g of `CH_(4)` are taken. For these gases:A. `V_("rms")` values will be sameB. Kinetic energy per mol will be sameC. Total kinetic energy will sameD. Pressure will be same |
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Answer» Correct Answer - b (b) `KE=(3)/(2)RT` `U_("rms")=sqrt((3RT)/(M)` |
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| 243. |
At what temperature will the total KE of 0.3 mo of He be the same as the total KE of 0.40 mol of Ar at 400K?A. 533KB. 400KC. 346KD. 300K |
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Answer» Correct Answer - a |
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| 244. |
At what temperature will the total `KE` of `0.3 mol` of `He` be the same as the total `KE` of `0.40 mol` of `Ar` at `400K` ?A. `533K`B. `400K`C. `346K`D. `300K` |
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Answer» Correct Answer - 1 `[(3)/(2)nRT]_(He)=(3)/(2)nRT` `0.3T=0.4xx400` `T=533K` |
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| 245. |
`1` of `NO_(2)` and `7//8 L` of `O_(2)` at the same temperature and pressure were mixed together. What is the relation between the mases of the two gases in the mixture?A. `M_(N_(3))=3M_(O_(2))`B. `M_(N_(2))=8M_(O_(2))`C. `M_(N_(2))=M_(O_(2))`D. `M_(N_(2))=16M_(O_(2))` |
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Answer» Correct Answer - c |
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| 246. |
Which has maximum internal energy at `290K` ?A. Neon gasB. Nitrogen gasC. Ozone gasD. Equal |
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Answer» Correct Answer - 3 Internal energy of a gas `=(f)/(2)nRT` Internal energy of a gas `prop f " " (f=5` for ozone ) |
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| 247. |
Assuming that `O_(2)` molecule is spherical in shape with radiusw `2overset(@)A`, the percentage of the volume of `O_(2)` molecules to the total volume of gas at S.T.P. is :A. 0.0009B. 0.009C. 9.0E-5D. 0.00045 |
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Answer» Correct Answer - a |
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| 248. |
If pressure becomes double at the same absolute temperature on `2L CO_(2)`, then the volume of `CO_(2)` becomesA. 2 litresB. 4 litresC. 5 litresD. 7 litres |
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Answer» Correct Answer - A `V_(2)=(P_(1)V_(1))/(T_(1)).(T_(2))/(P_(2))=(P)/(2P)xx2l txx(2T)/(T)=2l t` |
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| 249. |
Which gas has a density of 0.71 `gL^(-1)` at `0^(@)C` and 1 atm?A. ArB. NeC. COD. `CH_(4)` |
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Answer» Correct Answer - d |
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| 250. |
The root mean square speed of an ideal gas at constant pressure varies with density d as .A. `d^(2)`B. `d`C. `sqrtd`D. `(1)/sqrtd` |
| Answer» `u_(rms) =sqrt((3PV)/(M))=sqrt((3P)/(d))` . | |